2009/6/18 PJ <af.gour...@videotron.ca>:
> Martin Scotta wrote:
>> It is a sintax error
>>
>> if (in_array($ex, $selected) <--- missing )
>> echo "<br />yes";
>> else echo "<br />no";
>>
>> On Thu, Jun 18, 2009 at 10:13 AM, PJ <af.gour...@videotron.ca
>> <mailto:af.gour...@videotron.ca>> wrote:
>>
>> Ford, Mike wrote:
>> > On 17 June 2009 14:30, PJ advised:
>> >
>> >
>> >
>> >> For the moment, I am trying to resolve the problem of
>> >> extracting a value
>> >> from a string returned by a query. I thought that in_array()
>> would do
>> >> it, but the tests I have run on it are 100% negative. The only
>> thing I
>> >> have not used in the tests is third parameter bool $strict
>> which only
>> >> affects case-sensitivity if the $needle is a string.
>> >>
>> >
>> > $strict has nothing whatsoever at all in any way to do with case
>> > sensitivity -- $strict controls whether the values are compared
>> using
>> > equality (==) or identity (===) tests. As is stated quite
>> clearly on the
>> > manual page, in_array() is always case-sensitive.
>> >
>> >
>> >> This leads me to
>> >> believe that in_array() is either inappropriately defined in
>> >> the manual
>> >> and not effective on associative arrays
>> >>
>> >
>> > Complete rubbish -- the in_array() manual page is excellent and
>> totally
>> > accurate, complete with 3 working examples.
>> >
>> >
>> I would really like to understand why my attempts to reproduce the
>> first
>> example just did not work.
>> Note also that the examples do not show in_array($string, $array)
>> My array was Array ([0]=>6[1]=>14....), so when I tried if
>> (in_array($string, $array) , echo $string did not return 14 as I had
>> expected; neither did if(in_array("14", $array) ... nor
>> if(in_array(14,
>> $array). It still does not... actually, the screen goes blank.
>> So, what am I doing wrong?
>> Here's what is not working... I'm trying to reproduce the example from
>> php.net <http://php.net>:
>>
>> $selected = array();
>> if ( ( $results = mysql_query($sql, $db) ) ) {
>> while ( $row = mysql_fetch_assoc($results) ) {
>> $selected[] = $row['id'];
>> }
>> }
>> print_r($selected);
>> $ex = 14;
>> if (in_array($ex, $selected)
>> echo "<br />yes";
>> else echo "<br />no";
>>
>> Regardless if I put 14 into $ex or "14" or '14' or even if I put
>> the 14
>> instead of the $ex into the if line, I get a blank screen. It seems tp
>> me, from what I see in the manual that this should work... or am I
>> supposed to know something that is not clear in the examples... ?
>>
>>
>> --
>> Hervé Kempf: "Pour sauver la planète, sortez du capitalisme."
>> -------------------------------------------------------------
>> Phil Jourdan --- p...@ptahhotep.com <mailto:p...@ptahhotep.com>
>> http://www.ptahhotep.com
>> http://www.chiccantine.com/andypantry.php
>>
>>
>> --
>> PHP General Mailing List (http://www.php.net/)
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>>
>>
>>
>>
>> --
>> Martin Scotta
> Actually, we have to call it a typo because i fixed that and the results
> are exactly the same = blank screen. :-(

Advertising

If you're getting a blank screen instead of an error message I highly
recommend turning display_errors on and setting error_reporting to
E_ALL in you dev php.ini - that way you'll actually see error messages
rather than having to hunt through the code by hand.
-Stuart
--
http://stut.net/
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